Divide (a² − 3a − 18)/(2a² − 11a − 6) by (a² + a − 6)/(2a² − a − 1) and simplify.

Step-by-step solution

1. 1

   Factor the two monic quadratics. For $a^{2}-3a-18$ find numbers with product $-18$ and sum $-3$; for $a^{2}+a-6$, product $-6$ and sum $1$:

   $a^{2}-3a-18=(a-6)(a+3),\qquad a^{2}+a-6=(a+3)(a-2)$

2. 2

   Factor the two non-monic quadratics with the ac method. For $2a^{2}-11a-6$: $ac=-12$, and $-12$ and $1$ sum to $-11$; for $2a^{2}-a-1$: $ac=-2$, and $-2$ and $1$ sum to $-1$:

   $2a^{2}-11a-6=(2a+1)(a-6),\qquad 2a^{2}-a-1=(2a+1)(a-1)$

3. 3

   Invert the divisor and write everything factored.

   $\frac{(a-6)(a+3)}{(2a+1)(a-6)}\cdot\frac{(2a+1)(a-1)}{(a+3)(a-2)}$

4. 4

   Cancel across the product. In a product of fractions any top factor may cancel with any bottom factor: $(a-6)$, $(2a+1)$ and $(a+3)$ each appear once above and once below:

   $\frac{a-1}{a-2}$

   $\boxed{\dfrac{a-1}{a-2}}\qquad\left(a\neq 6,\ -\tfrac12,\ -3,\ 2\right)$

5. 5

   Check with a test value. At $a=0$: the original is $\dfrac{-18}{-6}\div\dfrac{-6}{-1}=3\div 6=\tfrac12$, and the answer gives $\dfrac{-1}{-2}=\tfrac12$ ✓.

6. 6

   Record all the excluded values. Only $a=2$ is visible in the final form, but $a=6$, $a=-\tfrac12$ and $a=-3$ were also excluded by the original expression's denominators and by the divisor being nonzero — they must be carried along even though the simplified fraction is perfectly well defined there.